Does anyone know of sizing guidelines or calculators for solar powering electronics? i.e. given power consumption, latitude and allowing for bad weather, accumulated dust and bird poop etc, calculate the size of solar cell and amount of storage to ensure that the electronics never run out of power?


Never is a long time ! :-)

Obtain sunshine hours per typical day on a typical month at your location from the superb gaisma insolation et al site here. As you are in NZ Ive chosen the Wellington page.

Th 4th graph down says the monthly mean insolation in kWh/m²/day Jan to Dec is as below. This is the equivalent full sunshine hours. Lowest is 1.4 hours/day in 6th entry = June


1.4 hours per day means over a 24 hour period you will get a mean solar Wattage per square metre of

  • Watts = 1000 x 1.4/24 = 58 Watts per square meter of isolation = 5.8% of full sun panel power

Solar panel efficiency = Zp typically ~ 13%. Bird poop degradation factor = Kbp = depends on cleaning etc.
Say Kbp= 0.75.

Average Wellinton June day = 1.4 sunshine hour/day but some days Watts ~= 0
So D = days you want to run with NO solar input.

I'm going to stop naming K factors and lump them all into "Kother".
Kother is a degradation factor comprised of all the factors you can think of MULTIPLIED together.

Battery charge to discharge energy return - say 80% - depends on chemistry and several other factors.

Temperature - about 90%of rated at 25C as panel gets hot.

Panel matching to battery - addressed by eg MPPT controllers - Panel will be 18V oc for a 12V battery and energy loss without an MPPT controller will vary - say 80%.


A Panel rated at say 100 Watts will power equipment run 24/7 in June in Wellington with a Wattage of

  • 100 x 1.4 hrs/24 hrs x 0.75 bp x Kother (0.8 battery x 0.9 temperature x ...) / D days

    = 100 x 0.058 x 0.75 x 0.8 x 0.9 x ... =~ 3 Watts / D

Below, Zp is allowed for by using the panel rated power.

ie a 100 Watt panel will run 3 Watts of equipment in Wellington in June IF days are typical. If you get super black dark and stormy and want to last say 3 days with about no sun you get 3/3 days =1 Watt 24/7 per 100 W of panel with 3 days holdup.

Pretty stunning !

Battery sizing = 1 Watt Hour per Watt of load x 24 hrs/day x D days holdup / Kbattery to load.

Say a 1 Watt load and 3 days holdup = 1 x 24 x 3 / 0.75 say = ~= 100 Watt hours.

E&OE !!!!!!!!!!! - the battery size and panel size don't seem quite correct. I may have dropped a figure somewhere there BUT the principle should be obvious and straight forwards A major aspect is getting the degradation factors correct.

Ask questions if interested.

  • \$\begingroup\$ Thanks Russell. Yeah, I was hoping to avoid the figuring it out approach :) The "some days Watts ~= 0" seems pessimistic - any thoughts? Also needs to be some allowance for losses where the panel is not tracking the Sun. Just geometric for the incident light I guess. I did see one formula: 100W panel * 1.4/5.8 (winter/summer in Wgtn) * 1.5hrs/day avg peak sun in mid latitudes = 36Whr/day, plus Kother. Guess I need to make some measurements. \$\endgroup\$
    – John McC
    Dec 6 '11 at 1:58
  • \$\begingroup\$ What city are you in? Your 36 Wh/day tanslates to 1.5 Watt continuous and for Kother = 2/3 you get 1 Watt. Which is 3X my above result. I don't know why, but yu need to. \$\endgroup\$
    – Russell McMahon
    Dec 6 '11 at 2:36
  • \$\begingroup\$ That was a from a formula for sizing household solar power. Derate the panel by the ration of winter/summer insolation and make an assumption that you get that for 1.5hr/day at that power. So the 1.5hr is I guess a crude factor for cloud, not pointing at the Sun etc. I'm thinking I'll need to get a panel and make some measurements. \$\endgroup\$
    – John McC
    Dec 7 '11 at 0:31
  • \$\begingroup\$ What city / area? | Gaisma data should be good enough for actual mean insolation without fudge factors. | Other multipliers that I mention obiously need assessment in your situation | No, ~= 0is not a bad figure for some days. Panel output goes down reasonably linearly with light level. There will be some NZ days (especially as you go south) where some days are very dark and stormy all day. Well under 1.5 hours equivalent full sun would be possible occasionally even in Auckland. \$\endgroup\$
    – Russell McMahon
    Dec 7 '11 at 3:49
  • \$\begingroup\$ Christchurch. The other question is Sun angle. And easy enough geometric calculation, but do the panels turn light into power, equally from all directions? \$\endgroup\$
    – John McC
    Dec 8 '11 at 2:37

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